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9. (a)
f
(
x
)
5
}
e
x
1
e
x
x
2
} 5
1
1 }
x
e
2
x
}
f
9
(
x
)
5
}
2
xe
x
e
2
2
x
x
2
e
x
}5
}
2
x
e
2
x
x
2
}
}
2
x
e
2
x
x
2
} 5
0
x
(2
2
x
)
5
0
x
5
0 or
x
5
2
f
9
(
x
)
,
0 for
x
,
0 or
x
.
2
The graph decreases, increases, and then decreases.
f
(0)
5
1;
f
(2)
5
1
1 }
e
4
2
}
<
1.541
f
has a local maximum at
<
(2, 1.541) and has a local
minimum at (0, 1).
(b)
f
is increasing on [0, 2]
(c)
f
is decreasing on (
2‘
, 0] and [2,
‘
).
10.
f
(
x
)
5
}
x
1
x
sin
x
} 5
1
1 }
sin
x
x
}
,
x
±
0
Observe that
)
}
sin
x
x
}
)
,
1 since
)
sin
x
)
,
)
x
)
for
x
±
0.
lim
x
→
0
f
(
x
)
5
1
1
lim
x
→
0
}
sin
x
x
} 5
1
1
1
5
2
Thus the values of
f
get close to 2 as
x
gets close to 0, so
f
doesn’t have an absolute maximum value.
f
is not defined at
0.
Section 8.2 Exercises
1.
lim
x
→
‘
}
x
3
2
e
3
x
x
1
1
}5
lim
x
→
‘
}
3
x
2
e
2
x
3
} 5
lim
x
→
‘
}
6
e
x
x
} 5
lim
x
→
‘
}
e
6
x
} 5
0
x
3
2
3
x
1
1 grows slower than
e
x
as
x
→
‘
.
2.
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.
 Spring '08
 GERMAN

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